Question: Multiply the following complex numbers: $({2-i}) \cdot ({4+i})$
Answer: Complex numbers are multiplied like any two binomials. First use the distributive property: $ ({2-i}) \cdot ({4+i}) = $ $ ({2} \cdot {4}) + ({2} \cdot {1}i) + ({-1}i \cdot {4}) + ({-1}i \cdot {1}i) $ Then simplify the terms: $ (8) + (2i) + (-4i) + (-1 \cdot i^2) $ Imaginary unit multiples can be grouped together. $ 8 + (2 - 4)i - 1i^2 $ After we plug in $i^2 = -1$ , the result becomes $ 8 + (2 - 4)i - (-1) $ The result is simplified: $ (8 + 1) + (-2i) = 9-2i $